by [1]}. Therefore, we have
\(\sum _{\begin{array}{c}1\le {\mathbf {q}_1} \le {\mathbf {q}_2}\le Q \\ \mathbf {q}_1\parallel \mathbf {q}_2\end{array}} *{A(\mathbf {q}_1, B_{{\mathbf {q}_1}})\cap A(\mathbf {q}_2, B_{{\mathbf {q}_2}})} &\ll *{\sum _{q=1}^Q q{B_q}}^2 + \sum _{q=1}^Q{B_{q}} \frac{q^2}{\varphi (q)}\nonumber \\&\ll *{\sum _{{\mathbf {q}}=1}^Q {A(\mathbf {q}, B_{{\mathbf {q}}})}}^2 + \sum _{{\mathbf {q}}=1}^Q{A(\mathbf {q}, B_{{\mathbf {q}}})} \frac{{\mathbf {q}}}{\varphi ({\mathbf {q}})},\) 
